Question
Science Question on Newton’s Second Law Of Motion
A stone of 1 kg is thrown with a velocity of 20 m s–1 across the frozen surface of a lake and comes to rest after traveling a distance of 50 m. What is the force of friction between the stone and the ice?
Answer
Mass of stone m = 1 kg, initial velocity u = 20 ms-1, Final velocity v= O (Therefore, the stone comes to rest), distance covered s= 50 m.
From third equation of motion,
v2 = u2 + 2as
(0)2 = (20)2 + 2a(50)
100a=-400 a=-4ms-2
Here negative sign shows that there is retardation in the motion of the stone. Force of friction between stone and ice = Force required to stop the stone = ma = 1 x -4 = -4N OR 4N